Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
/*
Keep solving problems.
*/

import java.util.*;
import java.io.*;

public class CFA {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    final long MOD = 1000L * 1000L * 1000L + 7;
    int[] dx = {0, -1, 0, 1};
    int[] dy = {1, 0, -1, 0};

    void solve() throws IOException {
        int n = nextInt();
        int m = nextInt();
        int[][] mat = new int[n][m];
        for (int i = 0; i < n; i++) {
            mat[i] = nextIntArr(m);
        }

        boolean flag = true;
        if (m < n) {
            flag = false;
            mat = transpose(mat);
            int tmp = n;
            n = m;
            m = tmp;
        }

        List<String> res = new ArrayList<>();
        List<Integer> num = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int min = Integer.MAX_VALUE;
            for (int j = 0; j < m; j++) {
                min = Math.min(min, mat[i][j]);
            }

            for (int c = 0; c < min; c++) {
                if (flag) {
                    res.add("row");
                }
                else {
                    res.add("col");
                }
                num.add(i + 1);
            }
            for (int j = 0; j < m; j++) {
                mat[i][j] -= min;
            }
        }

        for (int j = 0; j < m; j++) {
            int min = Integer.MAX_VALUE;
            for (int i = 0; i < n; i++) {
                min = Math.min(min, mat[i][j]);
            }

            for (int c = 0; c < min; c++) {
                if (flag) {
                    res.add("col");
                }
                else {
                    res.add("row");
                }
                num.add(j + 1);
            }
            for (int i = 0; i < n; i++) {
                mat[i][j] -= min;
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (mat[i][j] != 0) {
                    out(-1);
                    return;
                }
            }
        }

        outln(res.size());
        for (int i = 0; i < res.size(); i++) {
            outln(res.get(i) + " " + num.get(i));
        }
    }

    int[][] transpose(int[][] mat) {
        int n = mat.length;
        int m = mat[0].length;
        int[][] res = new int[m][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                res[j][i] = mat[i][j];
            }
        }

        return res;
    }

    void shuffle(long[] a) {
        int n = a.length;
        for(int i = 0; i < n; i++) {
            int r = i + (int) (Math.random() * (n - i));
            long tmp = a[i];
            a[i] = a[r];
            a[r] = tmp;
        }
    }

    long gcd(long a, long b) {
        while(a != 0 && b != 0) {
            long c = b;
            b = a % b;
            a = c;
        }
        return a + b;
    }

    private void outln(Object o) {
        out.println(o);
    }
    private void out(Object o) {
        out.print(o);
    }
    public CFA() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }
    public static void main(String[] args) throws IOException {
        new CFA();
    }

    public long[] nextLongArr(int n) throws IOException{
        long[] res = new long[n];
        for(int i = 0; i < n; i++)
            res[i] = nextLong();
        return res;
    }
    public int[] nextIntArr(int n) throws IOException {
        int[] res = new int[n];
        for(int i = 0; i < n; i++)
            res[i] = nextInt();
        return res;
    }
    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }
    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}